The odds of being dealt a two-card hand of AA is 4/52 * 3/51, or about
221. What are the chances (in percentage) of being dealt 6 or more
hands of AA in one, 150-hand session?

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Clarification of Question by dzatkovich-ga on 02 Jun 2005 11:33 PDT

No, you read it wrong. I understand that each chance is independent of
the time after that. The chances of getting one AA in 150 hands is
150/221. However, what are the chances of getting SIX AA's in 150
hands? Substantially lower.

Hi, dzatkovich-ga:

The chance of getting 6 or more pairs of Ace's in 150 random two-card
hands (from a standard card deck) is 1 minus the chance of getting 5
or fewer such hands.

The usual way to calculate this is then to use the binomial
distribution for independent repeated trials (Bernoulli trials).  For
a single hand as you know:

  Pr( hand is AA ) = (4/52)*(3/51) = (1/13)*(1/17) = 1/221

  Pr( hand not AA ) = 1 - Pr( hand is AA ) = 220/221

Now compute the first six terms of the binomial expansion:

  ( Pr( hand is not AA ) + Pr( hand is AA ) )^150  =  1

and these terms correspond exactly to:

  Pr( 0 of 150 hands are AA ) = (220/221)^150

                              = 0.50647856561905368169250739768815

  Pr( 1 of 150 hands are AA ) = 150 * (220/221)^149 * (1/221)

                              = 0.34532629474026387388125504387829

  Pr( 2 of 150 hands are AA ) = C(150,2) * (220/221)^148 * (1/221)^2

                              = 0.11694004071886208456433409440424

  Pr( 3 of 150 hands are AA ) = C(150,3) * (220/221)^147 * (1/221)^3

                              = 0.026222918221805437144729463593678

  Pr( 4 of 150 hands are AA ) = C(150,4) * (220/221)^146 * (1/221)^4

                              = 0.0043804192938697718866763990321257

  Pr( 5 of 150 hands are AA ) = C(150,5) * (220/221)^145 * (1/221)^5

                              = 0.0005814011062772606322315947806276

Because these events are mutually exclusive, the probability of any
one occurring is the sum of all six probabilities:

  Pr( 5 or less of 150 hands are AA ) = 0.99992963970013210980173399

The complementary probability is then:

  Pr( 6 or more of 150 hands are AA ) = 0.000070360299867890198266

or roughly 7 times in 100,000.


regards, mathtalk-ga

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Clarification of Answer by mathtalk-ga on 02 Jun 2005 18:27 PDT

Since you specifically asked for a percentage:

  Pr( 6 or more hands of 150 are AA ) = 0.0070360299867890198266%

or approximately 7/1000 of 1%.

-- mathtalk-ga

Perfect. Exactly what I was looking for.

The answer here is simple, as long as the deck is complete for each
hand the chances are the same as it would be for one hand about 221.

The chances of flipping a coin and getting the same result is 50-50,
the chances of having it do that 100 times in a row is still 50-50.

When playing Roulette every spin has the same chance of getting black or red.